We'll work on the 1st equation. We notice that the bases of the multiplied factors are matching, so we'll add the exponents:

4^(x/y)*4^(y/x) = 4^(x/y + y/x) = 4^[(x^2 + y^2)/xy]

But 4 = 2^2 and we'll raise to the exponent [(x^2 + y^2)/xy], both sides:

4^[(x^2 + y^2)/xy] = 2^2*[(x^2 + y^2)/xy]

32 = 2^5

The 1st equation will become:

2^2*[(x^2 + y^2)/xy] = 2^5

Since the bases are matching, we'll apply one to one property:

2*[(x^2 + y^2)/xy] = 5

We'll note x/y = z => y/x = 1/z

2(z + 1/z) = 5

2z^2 - 5z + 2 = 0

We'll determine the roots:

z1 = [5+sqrt(25 - 16)]/4

z1 = (5+3)/4

z1 = 2

z2 = 1/2

We'll put x/y = z1 <=> x/y = 2 => x = 2y

x/y = 1/2 => y = 2x

We'll re-write the 2nd equation, using the product rule of logarithms:

log3 (x-y) + log3 (x+y) = 1

log3 (x^2 - y^2) = 1

We'll take antilogarithms:

x^2 - y^2 = 3

We'll put x = 2y:

4y^2 - y^2 = 3

3y^2 = 3

y^2 = 1

y1 = 1 => x = 2

y2 = -1 => x = -2

We'll put x = y/2:

y^2/4 - y^2 = 3

y^2 - 4y^2 = 12

-3y^2 = 12

y^2 = -4

y1 = 2i => x1 = 2i/2 = i

y2 = -2i => x = -i

Since x and y values have to be real for the logarithms to exist, we'll accept only the real solutions of the equation. Furthermore, the logarithms must be positive and the only solution that satifies all these requirements is (2, 1)